Abstract Algebra – Subgroup of Finite Group of Order 936


I'm very confuse with this exercise. The prime number decomposition of $|G|$ is $936=2^{3}3^{2}13$. That is, $G$ is not a $p$-group, has no order $pq$, $p^2q$, $p^mq$ or $p^mq^n$. And $|H|=117=3^{2}13$ is not a prime or power of prime. And $G$ is not a cyclic group. In other words, it seems to me that $G$ and $H$ do not fit any of the conditions and consequences of Sylow's theorems and their corollaries. Any tips on what to do? Is there a theorem that fits these conditions that I have missed?

Best Answer

By the Sylow Theorems, the number of $13$-Sylow subgroups must divide $|G|$ and be congruent to $1$ modulo $13$. So it is a divisor of $2^3\times 3^2$ that is congruent to $1$ modulo $13$.

You have possible values $1$, $2$, $4$, $8$, $3$, $6$, $12$, $24$, $9$, $18$, $36$, and $72$. Of these, the only one that is congruent to $1$ modulo $13$ is $1$. So $G$ has a unique, hence normal, subgroup of order $13$, $P$.

Since $P$ is normal, for every subgroup $K$ of $G$, we have $KP=PK$, hence $KP$ is always a subgroup of $G$. The order is $|KP| = |K||P|/|K\cap P|$. Since $|P|=13$, all you need is a subgroup of order $9$. Does $G$ have a subgroup of order $9$?

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